Fig. 4.4 shows the elevation of a truncated cylinder at both ends. Draw full size the given view. Produce the pattern development.

The length of the pattern development will be the circumference of the cylinder.

We use the 12 division points from the plan view. Let's assume point 0 is at the lowest part of the truncation (leftmost in elevation) and point 6 is at the highest part (rightmost in elevation). The horizontal distance $x$ from the lowest point of the truncation to a generator at an angle $\alpha$ (measured from the lowest point in the plan view projection) is given by $x = R(1 - \cos \alpha)$, where $R = D/2 = 25$ mm.
The height $h(\alpha)$ of each generator is $h(\alpha) = H_{min} + x \tan(\theta)$.

• $H_{min} \approx 51.13$ mm
• $\tan(30^\circ) \approx 0.57735$

Let's calculate the heights for each of the 12 generator lines:

• Point 0 ($\alpha = 0^\circ$): $x = 25(1 - \cos 0^\circ) = 0$. $h_0 = 51.13 + 0 \times 0.57735 = \mathbf{51.13\, \text{mm}}$
• Point 1 ($\alpha = 30^\circ$): $x = 25(1 - \cos 30^\circ) = 25(1 - 0.866) = 3.35$ mm. $h_1 = 51.13 + 3.35 \times 0.57735 = 51.13 + 1.93 = \mathbf{53.06\, \text{mm}}$
• Point 2 ($\alpha = 60^\circ$): $x = 25(1 - \cos 60^\circ) = 25(1 - 0.5) = 12.5$ mm. $h_2 = 51.13 + 12.5 \times 0.57735 = 51.13 + 7.22 = \mathbf{58.35\, \text{mm}}$
• Point 3 ($\alpha = 90^\circ$): $x = 25(1 - \cos 90^\circ) = 25(1 - 0) = 25$ mm. $h_3 = 51.13 + 25 \times 0.57735 = 51.13 + 14.43 = \mathbf{65.56\, \text{mm}}$
• Point 4 ($\alpha = 120^\circ$): $x = 25(1 - \cos 120^\circ) = 25(1 - (-0.5)) = 37.5$ mm. $h_4 = 51.13 + 37.5 \times 0.57735 = 51.13 + 21.65 = \mathbf{72.78\, \text{mm}}$
• Point 5 ($\alpha = 150^\circ$): $x = 25(1 - \cos 150^\circ) = 25(1 - (-0.866)) = 46.65$ mm. $h_5 = 51.13 + 46.65 \times 0.57735 = 51.13 + 26.93 = \mathbf{78.06\, \text{mm}}$
• Point 6 ($\alpha = 180^\circ$): $x = 25(1 - \cos 180^\circ) = 25(1 - (-1)) = 50$ mm. $h_6 = 51.13 + 50 \times 0.57735 = 51.13 + 28.87 = \mathbf{80.00\, \text{mm}}$ (This is $H_{max}$)

The heights for the remaining points are symmetrical:

• $h_7 = h_5 = \mathbf{78.06\, \text{mm}}$
• $h_8 = h_4 = \mathbf{72.78\, \text{mm}}$
• $h_9 = h_3 = \mathbf{65.56\, \text{mm}}$
• $h_{10} = h_2 = \mathbf{58.35\, \text{mm}}$
• $h_{11} = h_1 = \mathbf{53.06\, \text{mm}}$
• $h_{12} = h_0 = \mathbf{51.13\, \text{mm}}$

• Draw a horizontal line representing the base of the pattern. Its length should be the circumference $C \approx 157.08$ mm.
• Divide this line into 12 equal segments. Each segment will have a length of $C/12 \approx 157.08 / 12 \approx 13.09$ mm. Label these points 0, 1, 2, ..., 11, 0 (or 12).
• At each of these division points, draw a vertical line corresponding to the calculated height for that generator:
  • At points 0 and 12, draw a vertical line of height 51.13 mm.
  • At points 1 and 11, draw a vertical line of height 53.06 mm.
  • At points 2 and 10, draw a vertical line of height 58.35 mm.
  • At points 3 and 9, draw a vertical line of height 65.56 mm.
  • At points 4 and 8, draw a vertical line of height 72.78 mm.
  • At points 5 and 7, draw a vertical line of height 78.06 mm.
  • At point 6, draw a vertical line of height 80.00 mm.
• Connect the top ends of these vertical lines with a smooth curve. This curve forms the truncated edge of the pattern.
• The top edge of the pattern development will be a straight horizontal line at a height of 80 mm, connecting the top of the $h_0$ and $h_{12}$ lines.
• The resulting shape is the pattern development of the truncated cylinder.